2 boxes with 1 applied force

[fahlim003]

Homework Statement

Box A
Mass=3.2kg
Force of Friction=2 Newtons

Box B
Mass=2.1kg
Force of Friction=1 Newton

Force applied 10.5 Newtons [horizontally].

http://img162.imageshack.us/img162/7697/f1qd1.png

Q. Find the magnitude of the force of contact between the two boxes.

Homework Equations

F=ma

The Attempt at a Solution

I need to find acceleration first since the Force of Friction of both boxes is not greater than the Force Applied.

Fnet=(mA+mB)a

(mA+mB)a=Fa-(FfA+FfB)
(3.2kg+2.1kg)a=10.5N-(2N+1N)
(5.3kg)a=10.5N-(3N)
a=10.5N-(3N)/5.3kg
a=7.5N/5.3kg
a=1.415m/s/s

Now, I tired many, many methods of solving for this from here on, including the example from this page:https://www.physicsforums.com/showthread.php?t=202002

However, the result seems still to be wrong.
I got 7.5N as the internal force both boxes act upon each other.

Any help is appreciated although not dire.

 

[alphysicist]

Hi fahlim003,

The Attempt at a Solution

I need to find acceleration first since the Force of Friction of both boxes is not greater than the Force Applied.

Fnet=(mA+mB)a

(mA+mB)a=Fa-(FfA+FfB)
(3.2kg+2.1kg)a=10.5N-(2N+1N)
(5.3kg)a=10.5N-(3N)
a=10.5N-(3N)/5.3kg
a=7.5N/5.3kg
a=1.415m/s/s

Now, I tired many, many methods of solving for this from here on, including the example from this page:https://www.physicsforums.com/showthread.php?t=202002

However, the result seems still to be wrong.
I got 7.5N as the internal force both boxes act upon each other.

Any help is appreciated although not dire.

I'm not getting the answer of 7.5N. Can you show how you got it?

 

[baba89]

I would approach this problem by first analyzing the given information and identifying any assumptions that need to be made. In this case, it is assumed that the force is applied horizontally, and that there is no other external force acting on the system.

Next, I would use the equation F=ma to calculate the acceleration of the system. From the given information, we know that the net force acting on the system is equal to the applied force minus the sum of the forces of friction. So, we can write the equation as:

Fnet = Fa - (FfA + FfB)

Where:
Fnet = net force
Fa = applied force
FfA = force of friction on Box A
FfB = force of friction on Box B

Plugging in the given values, we get:

Fnet = 10.5N - (2N + 1N)
Fnet = 7.5N

Now, we can use the equation F=ma to solve for acceleration:

7.5N = (3.2kg + 2.1kg)a
a = 7.5N/5.3kg
a = 1.415m/s/s

Once we have calculated the acceleration, we can use it to find the force of contact between the two boxes. This can be done using Newton's second law, which states that the net force on an object is equal to its mass multiplied by its acceleration. In this case, the net force is the force of contact between the two boxes, and the mass is the combined mass of the two boxes. So, we can write the equation as:

Fcontact = (mA + mB)a

Plugging in the values, we get:

Fcontact = (3.2kg + 2.1kg)(1.415m/s/s)
Fcontact = 9.5N

Therefore, the magnitude of the force of contact between the two boxes is 9.5 Newtons.